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The Young–Laplace equation gives a pressure excess $2\sigma /R$ inside a spherical droplet. Near coexistence, the common chemical-potential shift is $\mu \simeq \alpha \delta$, where $\alpha =f^{\prime \prime }({\varphi }_B)$, while changing phase changes the composition by approximately $2{\varphi }_B$. Balancing the capillary pressure against this thermodynamic shift gives the Gibbs--Thomson relationHence the compositions immediately outside and inside arewith

For an axisymmetric surface $r=a(z,t)$, twice the mean curvature with the outward-normal convention isWriting $a=a_0+\eta$ and retaining linear terms gives

At the unperturbed boundary $r=a_0$, the linearized kinematic boundary condition, tangential-stress condition, and Young–Laplace equation areFor a normal mode $e^{ikz+st}$ these become $u(a_0)=s\eta$, ${\sigma }_{rz}(a_0)=0$, and

The Papkovich–Neuber representation of body-force-free Stokes flow iswhere $\mathbf{\Phi }$ and $\chi$ are harmonic.

Let $x=kr$ and $E=e^{ikz+st}$. Substitution of the given radial vector potential and scalar potential givesUsing the modified Bessel function identity $(x{I}_1{)}^{\prime }=x{I}_0$, the tangential stress simplifies toThe stress-free condition at $r=a$ therefore yields

Let $d(z,t)$ be the local water-film thickness. To first order in the interface amplitudes,The lubrication theory flux down the vertical surface, with a no-slip boundary condition at the ice and a stress-free boundary condition at the water-air interface, isFor the unperturbed film $p_z=0$, so

The linearized Young–Laplace equation gives the capillary-pressure perturbationBecause the prescribed volume flux is uniform, its first-order perturbation must vanish. Linearization of the flux law givesThus, with $\Gamma =\gamma /(3\rho g)$,and henceThe complex amplitude ratio records the phase shift caused by surface tension in the long-wave approximation.

For $a<0$ and $b>0$, planar coexistence occurs at $\varphi =\pm {\varphi }_B$ with ${\varphi }_B=\sqrt{-a/b}$. Put $\alpha =f^{\prime \prime }({\varphi }_B)=a+3b{\varphi }_B^2=-2a>0$. A common small shift $\delta$ gives equal chemical potentials to first order,The thermodynamic pressure is $P=\varphi \mu -f$. Its difference between the positive interior and negative exterior is thereforeFor a sphere the Young–Laplace equation gives $P_{+}-P_{-}=2\sigma /R$. Consequently the Gibbs--Thomson relation is

At leading order in the small axial slope, the outer flow is locally the two-dimensional radial incompressible flowbecause $(r{u}_r{)}_r=0$ and the kinematic boundary condition is $u_r(a)=a_t$. This field is harmonic as a vector field away from the axis, so the exterior pressure is spatially constant and may be set to zero. Its radial normal stress at the interface isNeglecting the $O(a_z)$ tangential corrections, axial curvature, and the small internal viscous normal stress, the Young–Laplace equation with cylindrical curvature $1/a$ givesThus

A symmetric thin viscous sheet with shear-free faces has leading longitudinal velocity independent of thickness coordinate $z$, and incompressibility gives $w=-z{u}_x$. Young–Laplace equation and the Newtonian fluid stress tensor give ${\sigma }_{zz}=-p_a+\gamma h_{xx}/2$ and ${\sigma }_{xx}=-p_a+\gamma h_{xx}/2+4\mu u_x$. The effective longitudinal tension per unit transverse width is $\mathcal{T}=4\mu h{u}_x+\gamma h{h}_{xx}/2-\gamma h_x^2/4+2\gamma$. The last two terms include the horizontal projections of the two surface tension pulls. Force balance gives ${\mathcal{T}}_x=0$, yielding the displayed longitudinal equation; the face kinematic conditions give the displayed conservation of mass equation.

For an ideal soap film with equal constant surface tension $\gamma$ on both faces, the effective tension is $\tau =2\gamma$. Choose a unit normal $N$ from the minus side to the plus side, and $\Delta p=p_{-}-p_{+}$. Translating the Young–Laplace equation to $S=-dN$ givesA single interface uses $\tau =\gamma$. An outward-oriented spherical soap bubble has $H=-1/R$, giving $\Delta p=4\gamma /R$. The positive-convex curvature sum in the Young–Laplace equation article is $-2H$. ^[ref]

An open film with the same resting gas on both sides has zero pressure jump. Neglecting film weight, soap film pressure balance gives $H=0$: a minimal surface. A pressurized bubble instead has nonzero mean curvature. These statements assume equilibrium, constant surface tension, and negligible normal viscous stress. Surfactant gradients or motion require the fuller normal stress balance. ^[ref]

The variational outline is short. With fixed boundary, a normal displacement $fN$ gives $\delta A=-2{\int }_{M}HfdA$ by the first variation of area formula. For a closed interface oriented outward, $\delta V={\int }_{M}fdA$, so stationarity of $\tau A-\Delta pV$ gives $-2\tau H-\Delta p=0$. This is a first variation calculation; testing for a stable minimal surface needs the second variation, and deciding a global minimum needs a comparison of competitors.

Surface tension $\gamma$ is the tensile force per unit length across a line drawn in a liquid interface. For a clean interface with isotropy, it has the same value in every tangential direction: each side pulls the other along the surface, perpendicular to the line. A straight segment of length $\ell$ therefore transmits a force of magnitude $\gamma \ell$ from one interface. Here we assume constant $\gamma >0$, with fixed temperature and composition. The same coefficient measures the reversible work required to create surface area:The equality of units follows from $1\mathrm{J}=1\mathrm{N}\mathrm{m}$; the equality of the mechanical coefficients follows from the moving-rim argument below. ^{[ref] [ref]}

For a concrete example, take an ideal flat soap film spanning a rectangular wire frame with a straight sliding rim of length $\ell$. Pull the rim outwards in the plane of the film, perpendicular to the rim. Each of the film's two liquid–gas interfaces pulls the rim back towards the film with force $\gamma \ell$, so the outward holding force is $F=2\gamma \ell$. The two faces are two interfaces, each with its own surface tension. With the assumed toy value $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $\ell =20\mathrm{mm}=0.020\mathrm{m}$, the required force is $2.88\mathrm{mN}$; one interface would contribute $1.44\mathrm{mN}$. This value of $\gamma$ is a stipulated model parameter, rather than a measured value for a particular soap solution.

Why does an interface resist being enlarged? Its molecules have a different environment from those in the bulk liquid; changing that environment has an energy cost. In the constant-surface tension model, reducing surface area lowers surface energy. This connects an easily measured rim force to the shape of a droplet, the Young–Laplace equation for its pressure, and the restoring mechanism of capillary waves. It is useful because one coefficient links forces, energy and geometry. ^[ref]

The moving-rim argument also explains the factor of two without memorizing a special formula. An outward displacement $dx$ creates surface area $\ell dx$ on each face, hence $d{A}_{\mathrm{total}}=2\ell dx$. Slow pulling without viscous dissipation requires work $Fdx=\gamma d{A}_{\mathrm{total}}$, giving $F=2\gamma \ell$. For one interface, the same argument gives $F=\gamma \ell$. A surface tension gradient instead produces Marangoni stress; the constant-$\gamma$ assumption excludes that additional effect.

The Young–Laplace equation relates the pressure jump across a stationary interface to its curvature and surface tension. Choose a unit normal vector $\mathbf{n}$ pointing from the inside region to the outside region, and define $\kappa ={\nabla }_s\cdot \mathbf{n}$ to be positive for a convex sphere. In mechanical equilibrium, with constant surface tension and no additional normal viscous stresses,Here $k_1,k_2$ are the signed principal curvatures in this convention. Their sum is twice the corresponding signed mean curvature; stating the convention prevents a factor-of-two ambiguity. The Young–Laplace equation is a local normal stress balance for one interface. ^[ref]

A spherical liquid droplet of radius $R$ in a gas has one interface, with $k_1=k_2=1/R$. An ideal thin soap bubble has two liquid–gas interfaces. Adding their pressure jumps givesFor the soap bubble, $\Delta p$ now means the inner gas pressure minus the outer gas pressure. The second formula assumes equal surface tensions on both faces and film thickness negligible relative to $R$, so their radii can be identified. Retaining distinct radii gives $2\gamma /R_{\mathrm{inner}}+2\gamma /R_{\mathrm{outer}}$. A gas bubble inside bulk liquid has only one interface and therefore uses $2\gamma /R$. ^[ref]

With the assumed constant toy surface tension $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, a droplet with $R=1\mathrm{mm}$ has excess pressure $144\mathrm{Pa}$; at $R=0.5\mathrm{mm}$ it has $288\mathrm{Pa}$. A thin soap bubble assigned the same $\gamma$ and $R=1\mathrm{mm}$ has excess pressure $288\mathrm{Pa}$. These are differences from the surrounding pressure, rather than absolute pressures. Halving the radius doubles the capillary pressure; doubling the interface count also doubles it at fixed radius and surface tension.

The spherical case has a short work argument. A virtual increase in radius gives $dV=4\pi R^{2}dR$ and $dA=8\pi RdR$. At mechanical equilibrium, the net pressure work balances the increase in surface energy:For a thin soap bubble, both interfaces gain approximately the same surface area, doubling the right-hand side and giving $4\gamma /R$. This is a virtual work balance between neighboring spherical configurations, rather than an assumption that a fixed-volume droplet actually expands. The general Young–Laplace equation replaces the spherical area-to-volume ratio by local curvature, making geometry directly measurable through pressure.